## A nice little triangle puzzle

Here is a nice little puzzle I saw from brilliant.org on Facebook.

Have you worked it out yet?

Here’s what I did:

First I drew a diagram (obviously).

And worked out the area of the triangle.

Then the area of each sector.

## Circles puzzle

Here’s a lovely puzzle I saw on Brilliant.org this week:

It’s a nice little workout. I did it entirely in my head and that is my challenge to you. Do it, go on. Do it now….

Scroll down for my answer….

Have you done it? You better have…..

I looked at this picture and my frat thought was that the blue and gold areas are congruent. Thus the entire picture has an area of 70. There are 4 overlaps, each has an area of 5, so the total area of 5 circles is 90. Leaving each circle having an area of 18.

This is a nice mental work out and I feel it could build proprtional reasoning skills in my students. I am hoping to try it on some next week.

Did you manage the puzzle? Did you do it a different way?

*This post was cross posted to better questions *here*.*

## A lovely simple trigonometry puzzle

Sometimes a puzzle can look complicated, but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms, it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

*Sin^6 + sin^4 cos^2 – sin^4*

See it now? What if I rewrite it as:

*Sin^4 sin^2 + sin^4 cos^2 – sin^4*

I’m sure you have seen it now, but to be complete, take the common factor of the first two terms:

*Sin^4 (sin^2 + cos^2) – sin^4*

Obviously sin^2 + cos^2 = 1, so we’re left with:

*Sin^4 – sin^4 = 0*

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

*Cross-posted to Betterqs here. *

Sometimes a puzzle can look complicated, but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms, it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

*Sin^6 + sin^4 cos^2 – sin^4*

See it now? What if I rewrite it as:

*Sin^4 sin^2 + sin^4 cos^2 – sin^4*

I’m sure you have seen it now, but to be complete, take the common factor of the first two terms:

*Sin^4 (sin^2 + cos^2) – sin^4*

Obviously sin^2 + cos^2 = 1, so we’re left with:

*Sin^4 – sin^4 = 0*

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

## A lovely angle puzzle

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

*7x + 2y + 6z – 20 = 360*

*7x + 2y + 6z = 380 (1)*

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

*2x – 20 = 2y + 2z (2)*

*And*

*3x = 2x + 4z (3)*

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

*x = 4z (4)*

Subbing into 2 we get:

*8z – 20 = 2y + 2z*

*6z = 2y + 20 (5)*

Subbing into 1

*28z + 2y + 6z = 380*

*34z = 380 – 2y (6)*

Add equation (5) to (6)

*40z = 400*

*z = 10 (7)*

Then equation 4 gives:

*x = 40*

And equation 2 gives:

*60 = 2y + 20*

*40 = 2y*

*y = 20.*

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

*Cross-posted to Betterqs here.*

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December 9, 2015
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December 7, 2015
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## A puzzle with possibilities

Brilliant’s Facebook page is a fantastic source of brain teasers, they post a nice stream of questions that can provide a mental work out and that I feel can be utilised well to build problem solving amongst our students.

Today’s puzzle was this:

It’s a nice little question. But when I use it in class I will only use the graphic, as I feel the description gives away too much of the answer. Without the description students will need to deduce that the green area is a quarter of a circle radius 80 (so area 1600pi) with the blue semicircle radius 40 (so area 800pi) removed, leaving a green area of 800pi.

I find the fact that the area of the blue semi circle is equal to the green area is quite nice, and in feel that with a slight rephrasing the question could really make use of this relationship. Perhaps the other blue section could be removed or coloured differently and the question instead of finding the area could be find the ratio of blue area to green area.

Another option, one I may try with my further maths class on Friday, could be to remove the other blue section and remove the side length and ask them to prove that the areas are always equal, this would provide a great bit of practice at algebraic proof.

*Can you think of any further questions that could arise from this? I’d love to hear them!*

*This post was cross-posted to the blog Betterqs here.*

## Problem solving triangles

Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this one:

It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t, it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

*This post was cross posted to the BetterQs blog here.*